# 2019 AIME I Problems/Problem 7

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## Problem

There are positive integers $x$ and $y$ that satisfy the system of equations \begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*} Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$, and let $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$. Find $3m+2n$.

## Solution 1

Add the two equations to get that $\log x+\log y+2(\log(\gcd(x,y))+2(\log(\text{lcm}(x,y)))=630$. Then, we use the theorem $\log a+\log b=\log ab$ to get the equation, $\log (xy)+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630$. Using the theorem that $\gcd(x,y) \cdot \text{lcm}(x,y)=x\cdot y$, along with the previously mentioned theorem, we can get the equation $3\log(xy)=630$. This can easily be simplified to $\log(xy)=210$, or $xy = 10^{210}$.

$10^{210}$ can be factored into $2^{210} \cdot 5^{210}$, and $m+n$ equals to the sum of the exponents of $2$ and $5$, which is $210+210 = 420$. Multiply by two to get $2m +2n$, which is $840$. Then, use the first equation ($\log x + 2\log(\gcd(x,y)) = 60$) to show that $x$ has to have lower degrees of $2$ and $5$ than $y$ (you can also test when $x>y$, which is a contradiction to the restrains we set before). Therefore, $\gcd(x,y)=x$. Then, turn the equation into $3\log x = 60$, which yields $\log x = 20$, or $x = 10^{20}$. Factor this into $2^{20} \cdot 5^{20}$, and add the two 20's, resulting in $m$, which is $40$. Add $m$ to $2m + 2n$ (which is $840$) to get $40+840 = \boxed{880}$.

~minor mistake fix by virjoy2001 ~minor mistake fix by oralayhan

Remark: You can obtain the contradiction by using LTE. If $\nu_2{(x)}\geq{\nu_2{(y)}}, \nu_2{(y^2x)}=60$. However, $\nu_2{(xy)}=210$ a contradiction. Same goes with taking $\nu_5{(x,y)}$

## Solution 2

First simplifying the first and second equations, we get that

$$\log_{10}(x\cdot\text{gcd}(x,y)^2)=60$$ $$\log_{10}(y\cdot\text{lcm}(x,y)^2)=570$$

Thus, when the two equations are added, we have that $$\log_{10}(x\cdot y\cdot\text{gcd}^2\cdot\text{lcm}^2)=630$$ When simplified, this equals $$\log_{10}(x^3y^3)=630$$ so this means that $$x^3y^3=10^{630}$$ so $$xy=10^{210}.$$

Now, the following cannot be done on a proof contest but let's (intuitively) assume that $x and $x$ and $y$ are both powers of $10$. This means the first equation would simplify to $$x^3=10^{60}$$ and $$y^3=10^{570}.$$ Therefore, $x=10^{20}$ and $y=10^{190}$ and if we plug these values back, it works! $10^{20}$ has $20\cdot2=40$ total factors and $10^{190}$ has $190\cdot2=380$ so $$3\cdot 40 + 2\cdot 380 = \boxed{880}.$$

Please remember that you should only assume on these math contests because they are timed; this would technically not be a valid solution.

## Solution 3 (Easy Solution)

Let $x=10^a$ and $y=10^b$ and $a. Then the given equations become $3a=60$ and $3b=570$. Therefore, $x=10^{20}=2^{20}\cdot5^{20}$ and $y=10^{190}=2^{190}\cdot5^{190}$. Our answer is $3(20+20)+2(190+190)=\boxed{880}$.

## Solution 4

We will use the notation $(a, b)$ for $\gcd(a, b)$ and $[a, b]$ as $\text{lcm}(a, b)$. We can start with a similar way to Solution 1. We have, by logarithm properties, $\log_{10}{x}+\log_{10}{(x, y)^2}=60$ or $x(x, y)^2=10^{60}$. We can do something similar to the second equation and our two equations become $$x(x, y)^2=10^{60}$$ $$y[x, y]^2=10^{570}$$Adding the two equations gives us $xy(x, y)^2[x, y]^2=10^{630}$. Since we know that $(a, b)\cdot[a, b]=ab$, $x^3y^3=10^{630}$, or $xy=10^{210}$. We can express $x$ as $2^a5^b$ and $y$ as $2^c5^d$. Another way to express $(x, y)$ is now $2^{min(a, c)}5^{min(b, d)}$, and $[x, y]$ is now $2^{max(a, c)}5^{max(b, d)}$. We know that $x, and thus, $a, and $b. Our equations for $lcm$ and $gcd$ now become $$2^a5^b(2^a5^a)^2=10^{60}$$ or $a=b=20$. Doing the same for the $lcm$ equation, we have $c=d=190$, and $190+20=210$, which satisfies $xy=210$. Thus, $3m+2n=3(20+20)+2(190+190)=\boxed{880}$. ~awsomek

## Solution 5

Let $x=d\alpha, y=d\beta, (\alpha, \beta)=1$. Simplifying, $d^3\alpha=10^{60}, d^3\alpha^2\beta^3=10^{510} \implies \alpha\beta^3 = 10^{510}=2^{510} \cdot 5^{510}$. Notice that since $\alpha, \beta$ are coprime, and $\alpha < 5^{90}$(Prove it yourself !) , $\alpha=1, \beta = 10^{170}$. Hence, $x=10^{20}, y=10^{190}$ giving the answer $\boxed{880}$.

(Solution by Prabh1512)

## Solution 6 (Official MAA)

The two equations are equivalent to $x(\gcd(x,y))^2=10^{60}$ and $y(\operatorname{lcm}(x,y))^2=10^{570}$ respectively. Multiplying corresponding sides of the equations leads to $xy(\gcd(x,y)\operatorname{lcm}(x,y))^2=(xy)^3=10^{630}$, so $xy=10^{210}$. It follows that there are nonnegative integers $a,\,b,\,c,$ and $d$ such that $(x,y)=(2^a5^b,2^c5^d)$ with $a+c=b+d=210$. Furthermore, $$\frac{(\operatorname{lcm}(x,y))^2}{x}=\frac{y(\operatorname{lcm}(x,y))^2}{xy}=\frac{10^{570}}{10^{210}}=10^{360}.$$ Thus $\max(2a,2c)-a=\max(2b,2d)-b=360.$ Because neither $2a-a$ nor $2b-b$ can equal $360$ when $a+c=b+d=210,$ it follows that $2c-a=2d-b=360$. Hence $(a,b,c,d)=(20,20,190,190$, so the prime factorization of $x$ has $20+20=40$ prime factors, and the prime factorization of $y$ has $190+190=380$ prime factors. The requested sum is $3\cdot40+2\cdot380=880.$

## Solution 7

Add the two equations and use the fact that $\gcd\left(x,y\right)\cdot\mathrm{lcm}\left(x,y\right)=xy$ to find that $xy=10^{210}$. So let $x=2^a5^b$ and $y=2^{210-a}5^{210-b}$ for $0\leq a,b\leq210$. If $a\geq105$ then the exponent of $2$ in $x\cdot\gcd\left(x,y\right)^2=10^{60}$ is $a+2\left(210-a\right)=420-a$, so $a=360$, contradiction. So $a<105$. Then the exponent of $2$ in $x\cdot\gcd\left(x,y\right)^2$ is $a+2a=3a$, so $a=20$. Similarly, $b=20$. Then $3m+2n=3\left(a+b\right)+2\left(420-a-b\right)=\boxed{880}$ as desired.

~from trumpeter in the AoPS Forums Contest Discussion

## Video Solution(Pretty Straightforward)

https://www.youtube.com/watch?v=NOLk9-A4eDo Remember to subscribe!

~North America math Contest Go Go Go

## See Also

 2019 AIME I (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.